Answer
$$\frac{1}{{24}}\ln \left| {\sin 8{x^3}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\cot 8{x^3}} dx \cr
& {\text{set }}u = 8{x^3}{\text{ then }}\frac{{du}}{{dx}} = 24{x^2},\,\,\,\,\,\,\,\,\frac{{du}}{{24{x^2}}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{x^2}\cot 8{x^3}} dx = \int {{x^2}\cot u} \left( {\frac{{du}}{{24{x^2}}}} \right) \cr
& {\text{cancel the common factor}} \cr
& = \int {\cot u} \left( {\frac{{du}}{{24}}} \right) \cr
& {\text{use multiple constant rule}} \cr
& = \frac{1}{{24}}\int {\cot u} du \cr
& {\text{ using the Basic Trigonometric integral }} \cr
& \int {\cot x} dx = \ln \left| {\sin x} \right| + C\left( {{\text{see page 694}}} \right) \cr
& = \frac{1}{{24}}\ln \left| {\sin u} \right| + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{{24}}\ln \left| {\sin 8{x^3}} \right| + C \cr} $$
